1840 United States presidential election in Tennessee
Appearance
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Elections in Tennessee |
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A presidential election was held in Tennessee on November 3, 1840 as part of the 1840 United States presidential election.[1] Voters chose 15 representatives, or electors to the Electoral College, who voted for President and Vice President.
Tennessee voted for the Whig candidate, William Henry Harrison, over Democratic candidate Martin Van Buren. Harrison won Tennessee by a margin of 11.32%.
Conventions
[edit]Both William Henry Harrison and Martin Van Buren won their respective party conventions.
Results
[edit]United States presidential election in Tennessee, 1840[2] | ||||||||
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Party | Candidate | Running mate | Popular vote | Electoral vote | ||||
Count | % | Count | % | |||||
Whig | William Henry Harrison of Ohio | John Tyler of Virginia | 60,194 | 55.66% | 15 | 100.00% | ||
Democratic | Martin Van Buren of New York | Richard M. Johnson of Kentucky | 47,951 | 44.34% | 0 | 0.00% | ||
Total | 108,145 | 100.00% | 15 | 100.00% |
References
[edit]- ^ Dubin, Michael J. (2002). United States Presidential Elections, 1788–1860: The Official Results by County and State. Jefferson, NC: McFarland & Co. p. xvi.
- ^ "1840 Presidential General Election Results - Tennessee". U.S. Election Atlas. Retrieved December 23, 2013.